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Express the general solution of $$ z^{2}\left(z^{2}-1\right)^{2} y^{\prime \prime}+z\left(z^{2}-1\right)\left(2 z^{2}-1\right) y^{\prime}-\left[\left(3 \alpha^{2}-\frac{1}{4}\right) z^{2}+\alpha^{2}\right] y=0 $$ near \(z=1\) in terms of hypergeometric functions \({ }_{2} \boldsymbol{F}_{1}\)

Short Answer

Expert verified
Solutions are expressed using hypergeometric functions with configured parameters near the singular point.

Step by step solution

01

- Identify the type of differential equation

Observe that the given differential equation is a second-order linear differential equation with variable coefficients. It is suitable for expressing in terms of hypergeometric functions near the singular point.
02

- Transform the variable

Let’s perform a variable transformation to simplify the differential equation. Set \[ t = z - 1 \].Near \( z = 1 \), where \( t \to 0 \), and the differential equation becomes:\[ \frac{d^2 y}{dt^2} + \frac{4t + 2}{t(t + 2)} \frac{dy}{dt} - \frac{(3 \beta^2 - \frac{1}{4})(t+1)^2 + \beta^2}{t(t+2)^2} y = 0 \].
03

- Simplify the equation

For small t, we can approximate \[ (3 \beta^2 - \frac{1}{4})(t+1)^2 + \beta^2 \approx (3 \beta^2 - \frac{1}{4}) + (3 \beta^2 - \frac{1}{4}) 2t + \beta^2 = 3 \beta^2 t + \beta^2 - \frac{1}{4}.\] Near \( t = 0 \), the differential equation thus simplifies to:\[ t^2 (1 + t)^2 y'' + t (1 + t)(2 t + 1) y' - \big[(3\beta^2 - \frac{1}{4} t^2 + \beta^2 \big] y = 0 \].
04

- Solve using the hypergeometric function

Look for solutions of the form \[ y(t) = t^\rho \boldsymbol{F}(t) \] where \[ \boldsymbol{F}(t) \text{ is a hypergeometric function.} \] To match the equation with standard hypergeometric form, align terms with standard hypergeometric differential equation form:
05

- Determine parameter values

Compare the transformed equation with the general hypergeometric differential equation:\[ t(1 - t) y'' + [ c - (a + b + 1) t ] y' - ab y = 0 \]. This comparison provides parameters based on coefficients identified earlier.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

second-order linear differential equations
A second-order linear differential equation involves derivatives of a function up to the second order and has the general form:
Hypergeometric Functions and Series
Hypergeometric functions and series are a family of special functions defined by a series expansion. Hypergeometric series have the general form:

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